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How many three-digit positive integers are there with no repeated digits, if the first digit and the last digit must be odd, but the middle digit can be even or odd

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7 votes
Answer:
Answer is 160
Explanation:
The number of digits that exist is 10: 0,1,2,3,4,5,6,7,8,9
Number of all odd digits is 5 (1,3,5,7,9)
Number of all even digits is 5 (0,2,4,6,8)

We divide the problem into two cases: middle digit is even and middle digit is odd
Now to solve the problem list the digit possibilities for each case:

Case 1: middle digit is odd:
odd odd odd
Number of possibilities for hundreds digit: 5
Number of possibilities for tens digit: 4 (repetition not allowed)
Number of possibilities for unit digit: 3
Number of ways for case 1= 5x4x3=60

Case 2: middle integer is even:
odd even odd
Number of possibilities for hundreds digit : 5
Number of possibilities for tens digit: 5
Number of possibilities for unit digit: 4
Number of ways for case 2= 5x5x4=100

TOTAL number of ways= 60+100=160
HOPE THIS HELPS :)
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